Question

If four points are $A(6,3),B(-3,5),C(4,-2)$ and $P(x,y),$ then the ratio of the areas of $△PBC$ and $△ABC$ is:

• A. $\frac{x+y-2}{7}$
• B. $\frac{x-y-2}{7}$
• C. $\frac{x-y+2}{2}$
• D. $\frac{x+y+2}{2}$

Answer 1

The correct option is A $\frac{x+y-2}{7}$

Given: Coordinates of points $A(x_1,y_1)=(6,3),B(x_2,y_2)=(-3,5),C\left(x_{3,}{y}_3\right)=(4,-2)$ and $P(x,y).$

We know that the area of:

$△PBC=\frac{1}{2}[x(y_2-y_3)+x_3(y-y_2)+x_2(y_3-y)]$

$=\frac{1}{2}[x(5+2)+4(y-5)-3(-2-y)]$ $=\frac{1}{2}[7x+7y-14]$

Similarly, the area of

$△ABC=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)]+x_3(y_1-y_2)$

$=\frac{1}{2}[6(5+2)-3(-2-3)+4(3-5)]=\frac{49}{2}$

Therefore, the ratio of the areas of $△PAB$ and $△ABC$

$=\frac{7x+7y-14}{49}=\frac{7(x+y-2)}{49}=\frac{x+y-2}{7}$